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The speed ratio of a and B is 2:3, th two people walked a total of s km, then a's speed is (unit: km / h)
Speed of a = 2km
Because it has been made clear that a's speed is 2km, no matter how he goes, it is 2km
The walking speed ratio of a and B is 3:4. From a to B, B walked for 21 minutes. How many minutes does it take for a to walk?
To solve by proportion
Did you test the proportion of unit 3? So did I
It takes X minutes to set up a
3x=4×21
3x=84
x=28
A: A has to walk for 28 minutes
Must adopt me!!
The ratio of walking speed between a and B is 4:3. From a to B, a walked for 21 minutes. How many minutes did B walk?
The speed ratio is equal to the inverse of the time ratio
Therefore, B has to walk 21 △ 3 × 4 = 28 points
Class A and class B walk from the school along the same route to the military training site SKM away from the school to participate in the training. Half of the distance of class a walks at the speed of v1km / h
The other half of the distance is at the speed of v2km / h; the second half of the time is at the speed of v1km / h, and the time for class A and class B to walk to the military training base is t1h and t2h respectively
(1) The algebraic expression containing sv1.v2 is used to express t1h and t2h
(2) Please judge which of class A and B arrives at the military training base first? Why?
(1) A: T1 = s / (2v1) + S / (2v2) B: v1t2 / 2 + v2t2 / 2 = s, that is, T2 = 2S / (V1 + V2) (2) it is easy to know that both T1 and T2 are greater than 0t1 = s (V1 + V2) / (2v1v2), then T1 / T2 = (V1 + V2) &# 178; / (4v1v2) = (V1 & # 178; + V2 & # 178; + 2v1v2) / (4v1v2) = (V1 & # 178; + V2 & # 178;) / (4v1v2) + 1 / 2
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